Question 164257
Note: {{{a<>0}}} and {{{b<>0}}}


{{{(-5a^(-2)b^2)(-3a^3b^(-4))=(-5*-3)*(a^(-2)*a^3)*(b^2*b^(-4))=15a^(-2+3)*b^(2+(-4))=15a^1*b^(-2)=(15a)/(b^2)}}}


So {{{(-5a^(-2)b^2)(-3a^3b^(-4))=(15a)/(b^2)}}}